Early in the 19th century, a French geometer named Vecten studied a configuration of squares on the sides of a triangle. Vecten considered adding another layer of squares on the outer vertices of the Pythagorean squares, and then another one.

This configuration has many interesting properties, but Thanos Kalogerakis' message I mentioned at the outset, began with a statement, due to Hirotaka Ebisui. If we denote A=a^2,B=b^2,C=c^2, then the Pythagorean theorem appears simply as C=A+B.

I shall denote the next layer of Vecten's squares by A',B'C' and the next layer by A'',B''C'', as shown below:

Hirotaka Ebisui has found that in the case of the right triangle, A'+B'=5C'. On informing me of that result, Thanos added an identity for the next layer A''+B''=C''.

That was exciting enough for me to investigate. I can happily and proudly report that, for the next layer, A'''+B'''=5C'''.

Hence, I venture a general statement:

Let A,B,C be the Pythagorean squares on the sides of a right triangle so that A+B=C. Let's built recursively a sequence of triples of squares, A_k,B_k,C_k, with A_0=A,B_0=B,C_0=C, and A_{n+1},B_{n+1},C_{n+1} formed on the other vertices of (B_n,C_n),(C_n,A_n) and (A_n,B_n). Then
\begin{cases}A_n+B_n=C_n& n-even,\\A_n+B_n=5C_n& n-odd.\end{cases}